I am given that $Z$ is independent of $(X,U)$, where $Z$ and $X$ are binary random variables while $U$ is an arbitrary random variable. I need to prove the following: $$ H(X\oplus Z|U) \geq H(X|U)$$ Am I doing this right?: $$ H(X\oplus Z|U) - H(X\oplus Z|U,Z) + H(X\oplus Z|U,Z) - H(X|U)$$ $$ H(X\oplus Z|U) - H(X\oplus Z|U,Z) + H(X|U,Z) - H(X|U)$$ $$ H(X\oplus Z|U) - H(X\oplus Z|U,Z) + H(X|U) - H(X|U)$$ $$ I(X\oplus Z;Z|U) \geq 0$$ I manipulate the intial expression. I am not very confident about the proof. Can someone verify?

  • $\begingroup$ Check your expressions again. Should you been only having - and + in the first 3 lines? No inequalities or equivalences? $\endgroup$ Jun 24, 2020 at 14:18
  • $\begingroup$ @auspicious99 I was just playing with an expression. All 4 are equivalent, in the last line I could conclude that the expression is $\geq$ 0, which would then apply to the above 3 expressions as well. Sorry for the poor writing! $\endgroup$ Jun 24, 2020 at 14:33
  • $\begingroup$ @D.W. Yea, but I can't see how that helps! $\endgroup$ Jun 26, 2020 at 10:06
  • $\begingroup$ You're right, never mind my comment. Can you prove $H(X\oplus Z) \ge H(X)$? That might be easier to think about. $\endgroup$
    – D.W.
    Jun 26, 2020 at 17:38
  • $\begingroup$ @D.W. I am thinking along the same line, write $H(X)$ as $H(X\oplus Z|Z)$, it is clear then. $\endgroup$ Jun 27, 2020 at 15:28


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